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Question Number 127682 by Dwaipayan Shikari last updated on 31/Dec/20

(π^2 /(1+(π^2 /(3−π^2 +((9π^2 )/(5−3π^2 +((25π^2 )/(7−5π^2 +((49π^( 2) )/(9−7π^2 +((81π^2 )/(11−9π^2 +((121π^2 )/(.....))))))))))))))

$$\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\frac{\pi^{\mathrm{2}} }{\mathrm{3}−\pi^{\mathrm{2}} +\frac{\mathrm{9}\pi^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}\pi^{\mathrm{2}} +\frac{\mathrm{25}\pi^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}\pi^{\mathrm{2}} +\frac{\mathrm{49}\pi^{\:\mathrm{2}} }{\mathrm{9}−\mathrm{7}\pi^{\mathrm{2}} +\frac{\mathrm{81}\pi^{\mathrm{2}} }{\mathrm{11}−\mathrm{9}\pi^{\mathrm{2}} +\frac{\mathrm{121}\pi^{\mathrm{2}} }{.....}}}}}}} \\ $$

Commented by Dwaipayan Shikari last updated on 01/Jan/21

πtan^(−1) (π)

$$\pi{tan}^{−\mathrm{1}} \left(\pi\right) \\ $$

Commented by A8;15: last updated on 03/Jan/21

It will be nice if you show us a proof,sir!

$$\mathrm{It}\:\mathrm{will}\:\mathrm{be}\:\mathrm{nice}\:\mathrm{if}\:\mathrm{you}\:\mathrm{show}\:\mathrm{us}\:\mathrm{a}\:\mathrm{proof},\mathrm{sir}! \\ $$

Answered by Dwaipayan Shikari last updated on 03/Jan/21

tan^(−1) x=x−(x^3 /3)+(x^5 /5)−(x^7 /7)+.... = x+x(−(x^2 /3))+x(−(x^2 /3))(−((3x^2 )/5))+x(−(x^2 /3))(−((3x^2 )/5))(−((5x^2 )/7))+.. = (x/(1+((x^2 /3)/(1−(x^2 /3)+((x^2 /5)/(1−(x^2 /5)+((x^2 /7)/(1−(x^2 /7)+...))))))))=(x/(1+(x^2 /(3−x^2 +((3^2 x^2 )/(5−3x^2 +((5^2 x^2 )/(7−5x^2 +..)))))))) Take x=π

$${tan}^{−\mathrm{1}} {x}={x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}}−\frac{{x}^{\mathrm{7}} }{\mathrm{7}}+.... \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:{x}+{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)+{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)\left(−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}}\right)+{x}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}\right)\left(−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{5}}\right)\left(−\frac{\mathrm{5}{x}^{\mathrm{2}} }{\mathrm{7}}\right)+.. \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{{x}}{\mathrm{1}+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}}+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{5}}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{5}}+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{7}}}{\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{7}}+...}}}}=\frac{{x}}{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}−{x}^{\mathrm{2}} +\frac{\mathrm{3}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{5}−\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{5}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{7}−\mathrm{5}{x}^{\mathrm{2}} +..}}}} \\ $$$${Take}\:{x}=\pi \\ $$

Commented by Dwaipayan Shikari last updated on 03/Jan/21

https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula

Commented by A8;15: last updated on 03/Jan/21

Thanks sir. My respect for you!

$$ \\ $$$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{My}\:\mathrm{respect}\:\mathrm{for}\:\mathrm{you}! \\ $$

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